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borishaifa [10]
3 years ago
5

Explain how you would write 423,090,709,000 in word form

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0
Four hundred twenty-three billion, ninety million, seven hundred nine thousand


Hope this helped:))
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If what ? What is the If part, so I can help answer your question.
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8. Find all the real fourth roots of 256
mezya [45]
X^4=256
x^4-256=0
(x^2)^2-(16)^2=0
(x^2+16)(x^2-16)=0
either x^2+16=0
it gives complex roots.
or x^2-16=0
or x^2-4^2=0
(x+4)(x-4)=0
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7 0
3 years ago
What is the percent increase from 1 to 6
Advocard [28]

Answer: 20%

Step-by-step explanation:

1       5

100  x

100*5/1 = 20%

7 0
2 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
On the horizontal number line show how you will round 198.78 to the tens place & write the rounded number. Given picture is
harkovskaia [24]
If the three slashes are 190, 195, and 200, we can see it as 
<span>-l----------l--------|--l- with the slightly longer line being around 198.78. Since it is clearly closer to the end of the line than the start of the line, we round it to 200</span>
7 0
3 years ago
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